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ycow [4]
2 years ago
6

The half-life of a positron is very short. It reacts with an electron, and the masses of both are converted to two gamma-ray pho

tons. e+10 +0−1e⟶2???? This reaction is called an annihilation reaction. The mass of an electron or positron is 9.109×10−31 kg. (a) Calculate the energy produced by the reaction between one electron and one positron. (b) Assuming that the two γ-ray photons have the same frequency, calculate this frequency. J ????photon= Hz
Chemistry
1 answer:
sp2606 [1]2 years ago
5 0

Explanation:

(a)   It is known that relation between energy and mass is as follows.

            E = 2 \times mc^{2}

where,    E = energy

              m = mass

              c = speed of light = 3 \times 10^{8} m/s

As it is given that mass is 9.109 \times 10^{-31} kg. So, putting the given values into the above formula as follows.

             E = 2 \times mc^{2}

                       = 2 \times 9.109 \times 10^{-31} kg \times 3 \times 10^{8}m/s

                       = 1.638 \times 10^{-13} J

Therefore, we can conclude that the energy produced by the reaction between one electron and one positron is 1.638 \times 10^{-13} J.

(b) When gamma ray photons are produced then they will have the same frequency. Relation between energy and frequency is as follows.

                    E = h \times \nu   ..... (1)

where,     h = plank's constant = 6.626 \times 10^{-34} J.s

              \nu = frequency

Also,     E = 2 \times mc^{2} ........ (2)

Hence, equating equations (1) and (2) as follows.

                    h \times \nu = 2 \times mc^{2}        

So,    

6.626 \times 10^{-34} Js \times \nu = 1.638 \times 10^{-13} J

                           \nu = 1.236 \times 10^{20} Hz

Thus, we can conclude that the frequency is 1.236 \times 10^{20} Hz.

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What is the hydronium ion concentration of a solution with a pOH of 7.20?
-Dominant- [34]

Answer:

[H⁺] = 1.58 x 10⁻⁷ M.

Explanation:

∵ pOH = - log[OH⁻]

7.20 = - log[OH⁻]

log[OH⁻] = - 7.20

∴ [OH⁻] = 6.31 x 10⁻⁸.

∵ [H⁺][OH⁻] = 10⁻¹⁴.

∴ [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(6.31 x 10⁻⁸) = 1.585 x 10⁻⁷ M.

3 0
2 years ago
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Suppose that the microwave radiation has a wavelength of 12.4 cm . How many photons are required to heat 255 mL of coffee from 2
vivado [14]

Answer:

Explanation:

wavelength λ = 12.4 x 10⁻² m .

energy of one photon = h c / λ

= 6.6 x 10⁻³⁴ x 3 x 10⁸ /  12.4 x 10⁻²

= 1.6 x 10⁻²⁴ J .

Let density of coffee be equal to density of water .

mass of coffee = 255 x 1 = 255 g

heat required to heat up coffee = mass x specific heat x rise in temp

= 255 x 4.18 x ( 62-25 )

= 39438.3 J  .

No of photons required = heat energy required / energy of one photon

= 39438.3 / 1.6 x 10⁻²⁴

= 24649 x 10²⁴

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2 years ago
Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be
Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

                    F.C  =  +1

5 0
2 years ago
At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival
evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0
2 years ago
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Enter the balanced complete ionic equation for k2so4(aq)+cai2(aq)→caso4(s)+ki(aq). express your answer as a chemical equation. i
frozen [14]

The ionic equation is as below

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EXPLANATION

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ionic equation

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cancel the spectator ions that is 2k^+ and 2i^-

The net ionic equation is therefore

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2 years ago
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